Question

1, Thực hiện phép tính \(\dfrac{2xy}{x^2-y^2}+\dfrac{x-y}{2x+2y}+\dfrac{y}{y-x}\)

2, Phân tích đa thức thành nhân tử \(x^2-y^2-2y-1\)

Answer 1

1, \(\dfrac{2xy}{x^2-y^2}+\dfrac{x-y}{2x+2y}+\dfrac{y}{y-x}\)

\(=\dfrac{2xy}{x^2-y^2}+\dfrac{\left(x-y\right)\left(y-x\right)+2y\left(x+y\right)}{2\left(x+y\right)\left(y-x\right)}\)

\(=\dfrac{2xy}{x^2-y^2}+\dfrac{xy-x^2-y^2+xy+2xy+2y^2}{2\left(xy-x^2+y^2-xy\right)}\)

\(=\dfrac{2xy}{x^2-y^2}+\dfrac{4xy-x^2+y^2}{2\left(y^2-x^2\right)}\)

\(=\dfrac{2xy}{\left(x-y\right)\left(x+y\right)}+\dfrac{4xy-x^2+y^2}{2\left(y-x\right)\left(x+y\right)}\)

\(=\dfrac{-4xy}{\left(y-x\right)\left(y+x\right)}+\dfrac{4xy-x^2+y^2}{2\left(y-x\right)\left(x+y\right)}\)

\(=\dfrac{y^2-x^2}{2\left(y-x\right)\left(x+y\right)}=\dfrac{\left(y-x\right)\left(x+y\right)}{2\left(y-x\right)\left(x+y\right)}=\dfrac{1}{2}\)

2, \(x^2-y^2-2y-1\)

\(=x^2-\left(y^2+2y+1\right)\)

\(=x^2-\left(y+1\right)^2\)

\(=\left(x-y-1\right)\left(x+y+1\right)\)